You mean standardizing to the normal distributionZ=X−15.3 /4.2

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Similarly,E(X2)==02⋅p(0)+12⋅p(1)+22⋅p(2)+32⋅p(3)+42⋅p(4)+52⋅p(5)6.75,soV(X)=6.75−2.15=4.6andσ=4.6−−−√≈2.144.

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**Answer:**

**y=1.32x-0.0022**

**Step-by-step explanation:**

Let the standard equation of parabola be y=a+bx+c ……. (1)

**Let y be the altitude of the ball and x be the distance traveled by the ball.**

Now by putting the points one by one, we get the required equation.

first point = (0,0)

golf ball travels a distance of 600 feet along the ground. Therefore point = (600,0)

At altitude of 200, it travels the distance of 300 feet(mid point) = (300,200)

By putting y=0 and x=0 in (1),

we get c=0,

By putting y=0 and x=600 we get

0 = 360000a+600b …………..(2)

By putting y=200 and x=300, we get

200=90000a+300b …………….(3)

Solving (2) and(3) we get, a=-0.0022 and b = 1.32

**So the required equation is : **

**y= =0.0022+1.32x**

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You need to work out the variables of 20 (the last number) first:

1, 2, 4, 5, 10, 20

Lets call these folks “p”

At that point you need to work out the components of the main coefficient (the number before the most astounding force of x):

1

Lets call this person “q”

At that point you need to make all the conceivable divisions you can with:

±p/q

Since q is only 1, we are fortunate, so the conceivable normal roots are:

±1/1,±2/1,±4/1,±5/1.±10/1,±20/1

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